Application Of Derivatives

For this question, we have a polynomial function whose degree in n and it is given that,$ f(x) \frac{d^3 f}{d x^3}=n(n-1)(n-2) a_n x^{n-3}+\cdots \rightarrow =n-3$$ n \geq 3 n-3=n-5 n-3 n-5 \frac{d^3 f}{d x^3} \frac{d^5 f}{d x^5}(\mathrm{n}-2=0) f(x) $$ f(x)=2 x^2+x f(x) \Rightarrow f(5)=2(5)^2+5=55$

$ x=a g^{\prime \prime}(x) g^{\prime}(x) x=a a>0 g^{\prime \prime}(x)$ which may not necessarily be 0 , due to which we cannot comment if there is a point of inflection or not.

The first derivative: $ x=2 x=3 x<2( x=1): f^{\prime}(1)=6(1-2)(1-3)=6(-1)(-2)=12>0$$ $

To determine the number of distinct intersection points between the parabola and the lines defined by , we must consider the specific lines and .The critical point of the parabola can be found by differentiating its equation with respect to : Solving for , we have: The second derivative is constant: Since the second derivative is positive, this indicates a minimum point at .The minimum value of the parabola can be calculated by substituting into the equation of the parabola: This simplifies to: Since , the minimum value of is: Thus, the minimum value of the parabola is 2, which is above both lines and . Therefore, the parabola does not intersect either line, confirming that there are zero intersection points.

To understand the behavior of the function on the interval , we need to analyze how it behaves as compared to the basic sine function, .Understanding the Periodicity:The function has a period .For , the period is compressed to .Sine Function Behavior: is increasing on and decreasing on .Behavior of :Because the function's period is one-third that of , it will follow the increasing/decreasing pattern within each period segment.Evaluating Specific Points:At , .At , .At , .At , .Conclusion: is increasing from to . is decreasing from to .Thus, the function is increasing in the interval and decreasing in the interval .

Given that is a strictly decreasing function with for all , we define a function by .Since is strictly decreasing, its derivative is negative. So when we compute the derivative of , we have: Given the condition that , we find that is strictly positive, as cosine is positive when its argument is between and .Thus, combining these two facts: , and ,It follows that: Hence, the composition function is strictly decreasing on the interval .

As we can see from the figure, the area of any shape would be given by the equation:$ A(x), A^{\prime}(x)=0$$ A^{\prime \prime}(x) x=\frac{1}{\sqrt{3}}$$ $